Unformatted text preview: Using the Divergence Theorem we have integraldisplay R div F dV = integraldisplay S = âˆ‚R F Â· d S . Now the boundary of R , âˆ‚R = S , consists of 2 pieces: the circular disk at x = 5, S 1 , and the paraboloid, S 2 . We can parametrize S 1 by Î¦ 1 ( s, t ) = (5 , t cos s, t sin s ), 0 â‰¤ s â‰¤ 2 Ï€ , 0 â‰¤ t â‰¤ 2. This gives Ï† s Ã— Ï† t = ( t, , 0). We can parametrize S 2 by Î¦ 2 ( s, t ) = ( t 2 + 1 , t cos s, t sin s ), 0 â‰¤ s â‰¤ 2 Ï€ , 0 â‰¤ t â‰¤ 2. This gives Ï† s Ã— Ï† t = ( t, 2 t 2 cos s, 2 t 2 sin s ). For the Divergence Theorem we need outward normals, so for S 1 this means positive x â€“component and for S 2 negative x â€“component. Hence both Î¦ 1 and Î¦ 2 are orientation preserving. Now integraldisplay S F Â· d S = integraldisplay S 1 F Â· d S + integraldisplay S 2 F Â· d S = integraldisplay 2 Ï€ integraldisplay 2 (5 , t 2 cos 2 s, t sin s ) Â· ( t, , 0) dt ds + integraldisplay 2 Ï€ integraldisplay 2 ( t 2 +1 , t 2 cos 2 s, t sin s ) Â· ( t, 2 t 2 cos s, 2 t 2 sin s ) dt ds = integraldisplay 2 Ï€ integraldisplay 2 (2 t 4 cos 3 s + 2 t 3 sin 2 s t 3 + 4 t ) dt ds = integraldisplay 2 Ï€ parenleftbigg 2 6 5 cos 3 s + 2 3 sin 2 s 2 2 + 2 3 parenrightbigg ds = integraldisplay 2 Ï€ parenleftbigg 2 6 5 cos 3 s + 4(1 cos 2 s ) + 4 parenrightbigg ds = integraldisplay 2 Ï€ parenleftbigg 2 6 5 a24 a24 a24 a24 a58 = 0 cos 3 s a24 a24 a24 a24 a24 a58 = 0 4 cos 2 s + 8 parenrightbigg ds = 8(2 Ï€ ) = 16 Ï€ . We note that the two answers agree....
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 Winter '10
 EricMoore
 Multivariable Calculus, Cone

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